What is the most precise name for quadrilateral ABCD with vertices A(−1,0), B(1,2), C(4,2), and D(2,0)?

Question

A. rhombus
B. rectangle
C. parallelogram
D. square

Answer 1

 

SLOPES:

 

  AB = (2-0)/(1 - -1) = 2/2 = 1

  BC = (2-2)/(4-1) = 0/3 = 0

  CD = (2-0)/(4-2) = 2/2 = 1

  AD = (0-0)/(-1-2) = 0

 

So AB and CD are parallel and BC and AD are parallel because they have the same slope.

However, adjacent sides are NOT perpendicular

 

DISTANCES

 

AB = sqrt ( (2-0)^2 +(-1 -1)^2) = sqrt ( 2^2 + (-2)^2 ) = sqrt ( 4 + 4) = sqrt(8) = 2*sqrt(2)

 

BC = sqrt (  (2-2)^2 + (1-4)^2 ) = sqrt ( 0^2 + (-3)^2  ) = sqrt ( 0 + 9 ) = sqrt(9) = 3

 

CD = sqrt ( ( 2-0)^2 + (4-2)^2 ) = sqrt ( 2^2 + 2^2) = sqrt(4+4) = sqrt(8)  = 2*sqrt(2)

 

AD = sqrt ( (-1 - 2)^2 + (0-0)^2 ) = sqrt ( (-3)^2 + 0^2 ) = sqrt ( 9 + 0 ) = sqrt(9) = 3

 

opposite sides are congruent but adjacent sides are NOT

 

 

It cannot be a square NOR can it be a rhombus because all of the sides are NOT the same --> options D and A are eliminated

 

It cannot be a rectangle because adjacent sides are NOT perpendicular ---> option B is eliminated

 

 option C is the only answer: opposite sides are parallel and congruent; It is a parallelogram